Mathematics 100-201 Page 1 of 9 Student-No.:
Midterm 1 Duration: 50 minutes
This test has 6 questions on 9 pages , for a total of 40 points.
• Read all the questions carefully before starting to work.
• Q1 and Q2 are short-answer questions; put your answer in the boxes provided.
• All other questions are long-answer; you should give complete arguments and explanations for all your calculations; answers without justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Attempt to answer all questions for partial credit.
• This is a closed-book examination.
None of the following are allowed : documents, cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
First Name:
Student-No:
Last Name:
Section: 201
Signature:
Question: 1
Points: 9
Score:
2
12
3
4
4
4
5
6
6 Total
5 40
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1.
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2.
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3.
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4.
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5.
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Mathematics 100-201 Page 2 of 9 Student-No.:
9 marks
Short-Answer Questions.
Questions 1 and 2 are short-answer questions. Put your answer in the box provided. Full marks will be given for a correct answer placed in the box. Show your work also, for part marks. Each part is worth 3 marks, but not all parts are of equal difficulty.
Simplify your answers as much as possible in Questions 1 and 2.
1.
Determine whether each of the following limits exists, and find the value if they do. If a limit below does not exist, determine whether it “equals” ∞ , −∞ , or neither. You cannot use L’Hospital’s Rule (if you know what it is) for any of the limits.
(a) lim t → 1 t 2 − 6 t + 5 t 2 − 1
Answer: -2
Solution: Try direct substitution — get 0 / 0 so simplify.
t 2 − 6 t + 5 t 2 − 1
=
( t − 5)( t − 1)
( t − 1)( t + 1) t − 5
= t + 1 provided t = 1
Hence lim t → 1 t
2 − 6 t + 5 t 2 − 1
= lim t → 1 t − 5 t + 1
= − 2
(b) lim t → 2
1 t − 2
−
1
| t − 2 |
Answer: Does not exist
Solution: The left and right limits are different: lim t → 2 +
1 t − 2
−
1
| t − 2 |
= lim t → 2 +
1 t − 2
−
1 t − 2
= 0
1 lim t → 2
− t − 2
1
−
| t − 2 |
1
= lim t → 2
− t − 2
1
−
− ( t − 2)
2
= lim t → 2
− t − 2
= −∞
Thus the limit does not exist.
since since t t −
−
2
2
>
<
0
0
(c) lim x → + ∞
2 sin x + 4 x 3
Page 2

Mathematics 100-201 Page 3 of 9 Student-No.:
Answer: 0
Solution: sin x doesn’t have a limit as x → ∞ but it is at least bounded and the denominator goes to infinity. We will use the Squeeze Theorem .
− 1 ≤ sin x ≤ 1 ⇒ − 2 ≤ 2 sin x ≤ 2
⇒ − 2 + 4 ≤ 2 sin x + 4 ≤ 2 + 4
2
⇒ x 3
≤
2 sin x + 4 x 3
6
≤ x 3
Taking limits and employing the Squeeze theorem lim x → + ∞
2 sin x + 4 x 3
= 0
Page 3

Mathematics 100-201 Page 4 of 9 Student-No.:
12 marks 2.
(a) If f ( x ) = x
2
+9 x − 10 x
, x = 0, find f
0
( x ).
Answer:
Solution: f ( x ) = x 2
√ x
+
9 x
√ x
−
10
√ x
= x
3 / 2
+ 9 x
1 / 2 − 10 x
− 1 / 2 f
0
( x ) =
=
3 x
1 / 2
2
3 x
1 / 2
2
+
+
9 x
− 1 / 2
2
9 x
− 1 / 2
2
− (10 ·
− 1 x
− 3 / 2
)
2
+ 5 x
− 3 / 2
Alternatively, you can use the quotient rule.
(b) If g ( x ) =
1
2
( x 2 − 2 x ) e x , find g
0
( x ).
Answer:
Solution: Use the product rule.
( e x
)
0
= e x
(
1
2
( x
2 − 2 x ))
0
=
1
2
(2 x − 2)
= x − 1 g
0
( x ) = ( x − 1) e x
+
= e x
(( x − 1) +
1
2
1
2
( x
2
( x
2
− 2 x ) e x
− 2 x ))
(c) Let f ( x ) be a function differentiable at x = 2 and let g = x · f ( x ). The tangent line to the curve y = f ( x ) at x = 2 has slope 3 while the tangent line to the curve y = g ( x ) at x = 2 has slope 5. What is f (2)?
Answer: f (2) = − 1
Solution: We are told that f
0
(2) = 3 and g
0
(2) = 5.
g
0
( x ) = xf
0
( x ) + f ( x ) g
0
(2) = 2 f
0
(2) + f (2)
5 = 2 · 3 + f (2)
So f (2) = − 1.
(d) Let h ( x ) = f ( x )
− x x f ( x ) where f (2) = 1 , f
0
(2) = 3. Compute h
0
(2).
Page 4

Mathematics 100-201 Page 5 of 9 Student-No.:
Answer: h
0
(2) = 25 / 4
Solution: We use the quotient rule twice h
0 h
0
( x ) =
(2) =
= xf
0
( x ) − f ( x )
− x 2
2 · 3 − 1
5
4
−
2 2
− 5
1
−
=
1
25
4
−
1 2
2 · f (
3 x ) f
−
( x xf
) 2
0
( x )
Page 5

Mathematics 100-201 Page 6 of 9 Student-No.:
4 marks
Full-Solution Problems.
In questions 3–5, justify your answers and show all your work . If a box is provided, write your final answer there. Unless otherwise indicated, simplification of answers is not required in these questions .
3.
Show that the following equation has at least one solution: e x
− 2 cos x = 0
Solution: We use the Intermediate Value Theorem.
• We see that the function f ( x ) = e x − 2 cos x is the sum of two continuous functions so it is itself continuous (everywhere on the real line).
• At x = 0 , f (0) = 1 − 2 · 1 = − 1 < 0.
• At x = π/ 2: f ( π/ 2) = e
π/ 2 − 2 cos( π/ 2)
= e
π/ 2
+ 0
= e
π/ 2
> 0 , since e x
> 0 always
By the IVT since the function is continuous and is negative at x = 0 and positive at x = π/ 2, there exists some 0 < c
1
< π/ 2 so that f ( c
1
) = 0.
Page 6

Mathematics 100-201 Page 7 of 9 Student-No.:
4 marks 4.
Determine the horizontal asymptotes of the graph y = f ( x ) , where f ( x ) =
5 x 3 − 11 x 2 − 9 x
4 − 11 x − 2 x 3
Answer: y = tions
− 5
2 in both direc-
Solution: Try direct substitution — undetermined, so need more work. Factor out the highest power across
5 x 3 − 11 x 2 − 9 x
4 − 11 x − 2 x 3
=
= x x
3
3
(5 −
(
4 x
3
−
11 x
11
− x
2
9 x
2
)
− 2)
5 −
4 x
3
−
11 x
11
− x
2
9 x
2
− 2
Hence lim x → + ∞
5 x 3
4 −
− 11
11 x x 2
−
−
2 x
9
3 x
= lim x → + ∞
5
4 x
3
−
−
11 x
11
− x
2 x
9
2
− 2
= −
5
2
Also notice that lim x →−∞
5 x
3
4 −
− 11
11 x x
2
−
−
2 x
9
3 x
= lim x →−∞
5 x
4
3
−
−
11 x
11 x
2
− x
9
2
− 2
= −
5
2
Page 7

Mathematics 100-201 Page 8 of 9
6 marks 5.
Let a be a constant real number and let h ( x ) =
(
4 + 2 x
2
3 + 4 x − 2 x 2 x ≤ a x > a
For what values of a is h ( x ) continuous at x = a ?
Student-No.:
Solution:
• First h ( a ) = 4 + a 2
• Left-hand limit: lim x → a
− h ( x ) = lim x → a
−
4 + 2 x
2
= 4 + 2 a
2
= h ( a )
• Right-hand limit: lim x → a
+ h ( x ) = lim x → a
+
3 + 4 x − 2 x
2
= 3 + 4 a − 2 a
2
• So we must have
4 + 2 a
2
= 3 + 4 a − 2 a
2
4 a
2
− 4 a + 1 = 0
(2 a − 1)
2
= 0
So a = 1 / 2.
5 marks 6.
Let f ( x ) =
√
1 + 2 x
(a) What is the domain of this function?
Solution: We need 1 + 2 x ≥ 0 ⇔ x ≥ − 1
.
So the domain is [ − 1 / 2 , + ∞ )
2
(b) Find the derivative of f ( x ) at x = 0 using the definition . No points will be given for a different approach.
Page 8

Mathematics 100-201 Page 9 of 9 Student-No.:
Solution: 0 is in the domain of the function which is also continuous there.
f
0
(0) = lim h → 0
= lim h → 0
= lim h → 0 f (0 + h ) − f
√ h
1 + 2 h − 1
(0)
(
√ h
1 + 2 h − 1)( h (
√
1 + 2
√ h
1 + 2
+ 1) h + 1)
= lim h → 0 h (
1 + 2 h − 1
√
1 + 2 h + 1)
= lim h → 0
2 h h (
√
1 + 2 h + 1)
= lim h → 0
2
(
√
1 + 2 h + 1)
= 1
Page 9